Question

A 19.9 mL sample of a 1.76 M potassium sulfate solution is mixed with 14.4 mL of a 0.896 M barium nitrate solution and a precipitation reaction occurs. The precipitate is collected, dried, and found to have a mass of 2.54 g . Determine the percent yield of the reaction.
K2SO4(aq)+Ba(NO3)2(aq)→BaSO4(s)+2KNO3(aq)


The solid BaSO4 is collected, dried, and found to have a mass of 2.49 g . Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

Answer: The limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$     .....(1)

• For potassium sulfate:

Molarity of potassium sulfate solution = 1.76 M

Volume of solution = 19.9 mL

Putting values in equation 1, we get:

$1.76M=\frac{\text{Moles of potassium sulfate}\times 1000}{19.9}\\\\\text{Moles of potassium sulfate}=\frac{1.76\times 19.9}{1000}=0.035mol$

• For barium nitrate:

Molarity of barium nitrate solution = 0.896 M

Volume of solution = 14.4 mL

Putting values in equation 1, we get:

$0.896M=\frac{\text{Moles of barium nitrate}\times 1000}{14.4}\\\\\text{Moles of barium nitrate}=\frac{0.896\times 14.4}{1000}=0.013mol$

The chemical equation for the reaction of potassium sulfate and barium nitrate follows:

$K_2SO_4+Ba(NO_3)_2\rightarrow BaSO_4+2KNO_3$

By Stoichiometry of the reaction:

1 mole of barium nitrate reacts with 1 mole of potassium sulfate

So, 0.013 moles of barium nitrate will react with = $\frac{1}{1}\times 0.013=0.013mol$ of potassium sulfate

As, given amount of potassium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, barium nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 moles of barium nitrate produces 1 mole of barium sulfate.

So, 0.013 moles of barium nitrate will produce = $\frac{1}{1}\times 0.013=0.013moles$ of barium sulfate.

Now, calculating the mass of barium sulfate by using the equation:

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.013 moles

Putting values in above equation, we get:

$0.013mol=\frac{\text{Mass of barium sulfate}}{233.4g/mol}\\\\\text{Mass of barium sulfate}=(0.013mol\times 233.4g/mol)=3.03g$

• To calculate the percentage yield of barium sulfate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of barium sulfate = 2.49 g

Theoretical yield of barium sulfate = 3.03 g

Putting values in above equation, we get:

$\%\text{ yield of barium sulfate}=\frac{2.49g}{3.03g}\times 100\\\\\% \text{yield of barium sulfate}=82.18\%$

Hence, the limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.